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Chapter 4
BJT & FET
Frequency Response
Spring 2012
4th Semester Mechatronics
SZABIST, Karachi
CH 4
Frequency Response
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Course Support
Office: 100 Campus (404)
Official: ZABdesk
Subsidiary:
https://sites.google.com/site/zabistmechatronics/home/spring-2012/ecd
ebooks:
https://sites.google.com/site/zabistmechatronics/home/ebooks
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Frequency Response
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Chapter Contents
• BJT & JFET Frequency Response− Introduction
• Logarithms and Decibels
• General Frequency Considerations
• Bode plot − Low Frequency Analysis
• Low Frequency Response − BJT Amplifier
• Low Frequency Response − FET Amplifier
• High Frequency Response − BJT Amplifier
• High Frequency Response − FET Amplifier
• Multistage Frequency Effects*
3Frequency Response
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Introduction
Frequency Response
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Frequency Response:
Phase and amplitude plots and equations of an amplifier
Frequency Response Prerequisites:
1. Logarithms
2. Semi-log plots
3. Decibels
4. Normalization
Introduction 5
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Logarithms
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Logarithms:
The logarithm of a number is the exponent by which another fixed value, the base,
has to be raised to produce that number.
Common logarithms:
Natural logarithms:
Relationship of CL and NL:
Benefits:
• Plotting of a variable between wide limits
• Compression of large data
Logarithms 7
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,
2.3
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Logarithms:
Logarithms 8
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A nautilus displaying a
logarithmic spiral
Broccoli, which grows
in a logarithmic spiral
A low pressure area over
Iceland shows an
approximately logarithmic
spiral pattern
The whirlpool Galaxy
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Example 9-1:
Using the calculator, determine the logarithm of the following numbers to the base
indicated:
a. log10 106
b. loge e3
c. log10 10−2
d. loge e−1
Example 9-2:
Using the calculator, determine the logarithm of the following numbers:
a. log10 64
b. loge 64
c. log10 1600
d. log10 8000
Logarithms 9
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Example 9-3:
Using calculator, determine the antilogarithm of the following expressions:
a. 1.6 = log10 a
b. 0.04 = loge a
Example 9-4:
Using calculator, determine the logarithm of the following numbers:
a. log10 0.5
b. log10 (4000/250)
c. log10 (0.6 x 30)
Logarithms1
0
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1 0
1
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Semi−−−−logPlots
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Semilog Plots 12
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≅ 30%log102=0.3010
≅ 48%Log103 = 0.4771 log104 = 0.6021
(≅ 60%)
log10 9 = 0.9543
log10 8 = 0.9031
log10 7 = 0.8451
log10 6 = 0.7781
log10 5 = 0.6999
Linear
1
2
Semilog graph paper
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Semilog Plots 13
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Frequency Response
Identifying the numerical values of the tic marks on a log scale
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Example 9-5:
Determine the value of the point appearing on the logarithmic plot in Fig. 9-4 using
the measures made by a ruler (linear).
Semilog Plots 14
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10 10 ⁄
10 10
d1
d2
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Decibels
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Decibels 16
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! "#"$%
!& 10 "#"$'(%
!&) 10 **+,$-.)/% $'(0%
!1& 20 22 $'(%
|415 | =|41| ∙ |41| ∙ |417 |⋯⋯ |419|
!&5 = !&+ !&+ . . . . . + !&9
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Example 9-6:
Find the magnitude gain corresponding to a voltage gain of 100 dB.
Example 9-7:
The input power to a device is 10,000 W at a voltage of 1000 V. The output power is
500 W and the output impedance is 20 Ω.
Example 9-8:
An amplifier rated at 40 W output is connected to a 10 Ω speaker. Calculate:
a) The input power required for full power output if the power gain is 25 dB
b) The input voltage for rated output if the amplifier voltage gain is 40 dB
Decibels 17
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General Frequency
Considerations
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General Frequency Considerations:
The frequency response of an amplifier refers to the frequency range in
which the amplifier will operate with negligible effects from capacitors
and device internal capacitance.
This range of frequencies can be called the mid-range.
• At frequencies above and below the midrange, capacitance and any inductance
will affect the gain of the amplifier.
• At low frequencies the coupling and bypass capacitors lower the gain.
• At high frequencies stray capacitances associated with the active device lower the
gain.
• Also, cascading amplifiers limits the gain at high and low frequencies.
Freq. Considerations
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General Frequency Considerations:
Freq. Considerations
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• A Bode plot indicates the
frequency response of an
Amplifier:
• The horizontal scale
indicates the frequency (in
Hz) and the vertical scale
indicates the gain (in dB)
• The mid-range frequency
range of an amplifier is
called the bandwidth of the
amplifier
• The bandwidth is defined by
the lower and upper cutoff
frequencies
• Cutoff – any frequency at
which the gain has dropped
by dB
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General Frequency Considerations:
Freq. Considerations
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Normalization
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Normalization Process:
In communication, a decibel plot vs frequency is normally provided
rather than gain vs frequency
A process in which the vertical parameter is divided by a specific level or
quantity sensitive to a combination or variables of the system
The band frequencies define a level where the gain or quantity of interest
will be 70.7% or its maximum value
Normalization
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Normalization Process:
Normalization
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Normalized gain versus frequency plot
Decibels plots of the normalized gain versus frequency plot
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Example 9-9:Given the frequency response:
a) Find the cutoff frequency f1 and f2 using the measurements provided
b) Find the bandwidth of the response
c) Sketch the normalized response
Normalization
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Example 9-9:
Normalization
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dB Plot:
Normalization
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Decibel plot of the normalized gain versus frequency plot
Av/Avmid Av/Avmid|dB
1 0
0.707 -3
0.5 -6
0.35 -9
0.25 -12
Low Frequency Analysis
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Low Frequency RC Circuit Analysis:
LF Analysis−−−− Bode Plot
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Low frequency response for the R-C circuit
Low Frequency RC Circuit Analysis:
LF Analysis−−−− Bode Plot
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41 0.707|;<=,
> 1
2?@A41
11 B >>
41$CD% 20 1
1 >>
#
41$CD% 20 >> E≪E
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Low Frequency RC Circuit Analysis:
LF Analysis−−−− Bode Plot
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f f1/f Av(dB)
f1 1 0
½ f1 2 -6
¼ f1 4 -12
1/10 f1 10 -20
Low Frequency RC Circuit Analysis:
LF Analysis−−−− Bode Plot
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f f1/f Av(dB)
f1 1 0
½ f1 2 -6
¼ f1 4 -12
1/10 f1 10 -20
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Low Frequency RC Circuit Analysis:• The piecewise linear plot of the asymptotes and associated breakpoints is called a
Bode plot of the magnitude versus frequency
• A change in frequency by a factor of 2, equivalent to 1 octave, results in a
6-dB change in the ratio as noted by the change in gain from f1/2 to f1.
• For a 10:1 change in frequency, equivalent to one decade, there is a 20-dB change
in the ratio as noted by the change in gain from f1/10 to f1.
LF Analysis−−−− Bode Plot
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41 GH 10IJ$CD%/L M NOP >
>
Low Frequency RC Circuit Analysis:
LF Analysis−−−− Bode Plot
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Phase response for the RC circuit Example 9-9
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Example 9-10:
For the network of fig. 9-20: (R = 5 kΩ, C = 0.1 µF)
a) Determine the break frequency
b) Sketch the asymptotes and locate the −3 dB point
c) Sketch the frequency response curve
d) Find the gain at Av(dB) = − 6 dB
LF Analysis−−−− Bode Plot
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Example 9-10:
LF Analysis−−−− Bode Plot
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Example 9-10:
LF Analysis−−−− Bode Plot
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Computer Analysis
% bode plot of Example 9-10
f = 10:10^4;
fo = 318.5;
A = 20*log(1./(1+(fo./f).^2).^(1/2));
semilogx(f,A), xlabel('f (log scale)'),
ylabel('Av(dB)')
grid
101
102
103
104
-70
-60
-50
-40
-30
-20
-10
0
f (log scale)
Av(d
B)
Low Frequency Response
BJT amplifiers
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BJT Amplifiers:
.
LF Response −−−− BJT Amplifiers
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Effects of Cs on the LF response:
LF Response −−−− BJT Amplifiers
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Effects of CC on the LF response:
LF Response −−−− BJT Amplifiers
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Effects of CE on the LF response:
LF Response −−−− BJT Amplifiers
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Effects of Cs and CE on the LF response:
LF Response −−−− BJT Amplifiers
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LC
o L c
1f
2 ( )Cπ R R=
+
o C oR R ||r=
The cutoff frequency due to CC can
be calculated with
where
The cutoff frequency due to CS can be
calculated by
Ls
s i s
1f
2 (R R )Cπ=
+
i 1 2 eR R ||R ||βr=
where
Example 9-11:a) Determine the lower cutoff frequency for the network of Fig. 9.23 using the
following parameters:
CS = 10 µF, CE = 20 µF, CC = 1 µF,
RS = 1 kΩ, R1 = 40 kΩ, R2 = 10 kΩ, RE = 2 kΩ, RC = 4 kΩ, RL = 2.2 kΩ,
β = 100, ro = ∞ Ω, VCC = 20 V
a) Sketch the frequency response using a Bode plot
b) Verify the result using a Simulator.
LF Response −−−− BJT Amplifiers
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Example 9-11:
LF Response −−−− BJT Amplifiers
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Low Frequency Response
FET amplifiers
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FET Amplifiers:
LF Response −−−− FET Amplifiers
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FET Amplifiers:
LF Response −−−− FET Amplifiers
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sig i G
1
2 (R R )CLG
fπ
=+
i GR R=
The cutoff frequency due to
CG can be calculated with
where
o L G
1
2 (R R )CLC
fπ
=+
O D G||R R r=
The cutoff frequency due to
CC can be calculated with
where
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FET Amplifiers:
LF Response −−−− FET Amplifiers
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ReqLS
eq S
1
2 R Cf
π=
d
eq S
m Ω
1||
gr
R R
≅ ∞
=
The cutoff frequency due to
CS can be calculated with
where
Example 9-12:a) Determine the lower cutoff frequency for the network of Fig. 11.32 using the
following parameters:
CG = 0.01 F, CC = 0.5 F, CS = 2 F
Rsig = 10 k, RG = 1 M, RD = 4.7 k, RS = 1 k, RL = 2.2 k
IDSS = 8mA, VP= − 4 V rd = ∞ Ω , VDD = 20 V
b) Sketch the frequency response using a Bode plot.
LF Response −−−− FET Amplifiers
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Example 9-12:
LF Response −−−− FET Amplifiers
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High Frequency Response
FET amplifiers
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FET Amplifiers:
HF Response −−−− FET Amplifiers
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FET Amplifiers:
HF Response −−−− FET Amplifiers
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FET Amplifiers:
Capacitances that affect the high-frequency response are
• Junction capacitances
Cgs, Cgd, Cds
• Wiring capacitances
Cwi, Cwo
• Coupling capacitors
CG, CC
• Bypass capacitor
CS
HF Response −−−− FET Amplifiers
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FET Amplifiers:
HF Response −−−− FET Amplifiers
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Hi
Thi i
1
2 Cf
πR=
i Wi gs MiC C CC= + +
Mi v gd(1 A )CC = −
Thi sig GR R ||R= Ho
Tho o
1
2 R Cf
π=
o Wo ds MoC C CC = + +
Mo gd
v
11 C
AC
= −
Tho D L dR ||R ||rR =
Figure 9-64 (a) & (b)
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Example 9-14:
HF Response −−−− FET Amplifiers
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Square Wave
Testing
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Square Wave Testing:
Square Wave Testing
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Square Wave Testing:
Square Wave Testing
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Example 9-15:The application of a 1-mV, 5-kHz square wave to an amplifier resulted in the output
waveform of Fig. 9-72.
(a) Write the Fourier series expansion for the square wave through the ninth
harmonic.
(b) Determine the bandwidth of the amplifier
(c) Calculate the low cutoff frequency.
Square Wave Testing
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Reading:
1. Summary
2. Equations
3. Computer analysis
Problems:
1. Sec 8.2: (odd)
2. Sec 8.3: 17,18
3. Sec 8.4: 19,21
4. Sec 8.5:23,25
5. Sec 8.6: 27,29
6. Sec 8.7: 31
7. Sec 8.8: 33,35,37
8. Sec 8.10: 39,41
9. Sec 8.11: 43
10. Sec 8.12: 45
11. Sec 8.14: 47
12. Sec 8.15: 49
Home Task 62Frequency Response
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CH 1
References 63FET
1. Bolestad
2. Paynter
Frequency Response
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